SOIL MECHANICS (version Fall 2008) Presented by: Jerry Vandevelde, P. E. Chief Engineer GEM Engineering, Inc. 1762 Watterson Trail Louisville, Kentucky (502) 493-7100 1 National Council of Examiners for Engineering and Surveying http://www. ncees. org/ 2 STUDY REFERENCES • Foundation Engineering; Peck Hanson & Thornburn •Introductory Soil Mechanics and Foundations; Sowers •NAVFAC Design Manuals DM-7. 1 & 7. 2 •Foundation Analysis and Design; Bowles •Practical Foundation Engineering Handbook; Brown 3 Soil Classification Systems * Unified Soil Classification System * AASHTO Need: Particle Sizes and Atterberg Limits 4

Particle Sizes (Sieve Analysis) (Well Graded) (Poorly Graded) 0. 1 5 Atterberg Limits Liquid, Plastic & Shrinkage Limits Plasticity Index (PI) PI = Liquid Limit – Plastic Limit (range of moisture content over which soil is plastic or malleable) 6 UNIFIED SOIL CLASSIFICATION SYSTEM ASTM D-2487 7 8 Ref: Peck Hanson & Thornburn 2nd Ed. Effective Size = D10 10 percent of the sample is finer than this size D60 = 1. 6mm D30 = 0. 2mm D10 = 0. 03mm 0. 1 0. 1 9 Uniformity Coefficient (Cu) = D60/D10 Coefficient of Curvature (Cz) = (D30)2/(D10xD60) D60 = 1. 6mm D30 = 0. 2mm D10 = 0. 03mm 0. 1 10 Well Graded – Requirements 50% coarser than No. 00 sieve Uniformity Coefficient (Cu) D60/D10 >4 for Gravel > 6 for Sand Coefficient of Curvature (Cz) = (D30)2/(D10xD60) = 1 to 3 11 Is the better graded material a gravel? 81% Passing No. 4 18% Finer No. 200 0. 1 0. 1 12 Gravel if > 50 Percent Coarse Fraction retained on No. 4 sieve % Retained on No. 200 = 82% 1/2 = 41% 19% (100-81) retained on No. 4 sieve (gravel) 19< 41 half of coarse fraction 81% Passing No. 4 18% Finer No. 200 ? sand 0. 1 (“S”) 13 Well Graded Sand? Uniformity Coefficient (Cu) > 6 = D60/D10 Coefficient of Curvature (Cz) = 1 to 3 = (D30)2/(D10xD60) 14 D60 = 1. 6mm D30 = 0. 2mm D10 = 0. 3mm 0. 1 Well Graded Sand? Uniformity Coefficient (Cu) D60/D10 = 1. 6/. 03 = 53 > 6 D60 = 1. 6mm D30 = 0. 2mm D10 = 0. 03mm Coefficient of Curvature (Cz) = (D30)2/(D10xD60) = 0. 22/(. 03×1. 6) = 0. 83 12% Passing No. 200 sieve: GM, GC, SM, SC 0. 1 >12% passing No. 200 sieve Since = “S” ? SC or SM 16 What Unified Classification if LL= 45 & PI = 25? From sieve data SC or SM 0. 1 A) “SC” B) “SM” C) “CL” or D) “SC & SM” 17 Unified Classification Answer is “A” ? SC 18 AASHTO (American Association of State Highway and Transportation Officials) 19 What is the AASHTO Classification? 65% Passing No. 10 40% Passing No. 0 18% Finer No. 200 1) 18 % passing No. 200 sieve 2) 65% passing No. 10 sieve 3) 40% passing No. 40 sieve 4) assume LL = 45 & PI = 25 20 18 percent passing No. 200 sieve; 65 percent passing No. 10 sieve 40 percent passing No. 40 sieve; assume LL = 45 & PI = 25 21 AASHTO Classification 1 2 3 4 4 1) 18 % passing No. 200 sieve 2) 65% passing No. 10 sieve 3) 40% passing No. 40 sieve 4) assume LL = 45 & PI = 25 22 AASHTO Group Index 23 Mass-Volume (Phase Diagram) • Unit volume of soil contains: Total Volume Va Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil – Air (gases) – Water (fluid) – Solid Particles 24 Moisture Content = ? eight of water/ weight of dry soil ? = Ww/Wd water loss/(moist soil weight – water loss) ? = Ww/(Wm-Ww) and ? =(Wm-Wd)/Wd 25 Mass – Volume Relationships Density or Unit Weight = Moist Unit Weight = ? m ? ?m = Wm/Vt = ? d + ? ?d ? = (? m – ? d )/ ? d ? ?d + ? d = ? m ? m= (1+ ? ) ? d ? d = ?m/(1+ ? ) b 26 Total Volume = ? Volume (solid + water + air) = Vs+Vw+Va ? Va = Vt – Vs- Vw Total Volume Va Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 27 Relationship Between Mass & Volume Volume = Mass/(Specific Gravity x Unit Weight of Water) = Ws/(SGxWw) Va Total Volume Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 28

Specific Gravity = weight of material/ weight of same volume of water Soil Specific Gravity Typical Range 2. 65 to 2. 70 Specific Gravity of Water = 1 29 Saturation = S expressed as percent S = volume of water/ volume of voids x 100 Total Volume Va Air Total S = Vw/Vv x 100 Ww Ws Weight Vt Vv Vw Vs Water Wt Soil Always ? 100 30 Porosity n = volume of voids/ total volume n = Vv/Vt Void Ratio e = volume of voids/ volume of solids e = Vv/Vs Total Volume Va Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 31 What is the degree of saturation for a soil with: SG = 2. 68, ? m = 127. 2 pcf & ? = 18. 6 percent A) 88. 4 Total Volume Va

Air Total Vt Vv Vw Vs Water Ww Ws Weight B) 100. 0 Wt Soil C) 89. 1 32 What are the porosity and degree of saturation for a soil with: SG = 2. 68, ? m = 127. 2 pcf & ? = 18. 6 percent = 107. 3pcf ?d = ? m/(1+ ? ) = 127. 2/(1. 186) Total Volume Va Air Total Vt Vv Vw Vs Water Soil Ww Weight Wt Ws Ww = ? m- ? d = 19. 9 pcf Vw = Ww/62. 4 = 0. 319 cf Vs = ? d /(SGx62. 4) = 0. 642 cf Va = Vt – Vw – Vs = 1- 0. 319 – 0. 642 = 0. 039 cf Vv = Vw + Va = 0. 358 cf 33 What are the porosity and degree of saturation for a soil with: SG = 2. 68, ? m = 127. 2 pcf & ? = 18. 6 percent Vw = 0. 319 cf, Vs = 0. 642 cf, Vv = 0. 358 cf Total Volume

Va Air Total Degree of Saturation = Vw/Vv x 100 Ww Weight Wt Ws Vt Vv Vw Vs Water = 0. 319/0. 358 x 100 = 89. 1% Soil Answer is “C” 34 Ref: NAVFAC DM-7 35 Borrow Fill Adjustments Borrow Material Properties: ?m = 110 pcf & ? = 10% Placed Fill Properties: ? d = 105 pcf & ? = 20% How much borrow is needed to produce 30,000 cy of fill? How much water must be added or removed from each cf of fill? Total Volume Va Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 36 Borrow Fill Adjustments Borrow Material Properties: ?m = 110 pcf & ? = 10% ?d = ? m /(1+? ) = 110/(1. 10) =100 pcf; Ww = 110-100=10 lbs Placed Fill Properties: ? = 105 pcf & ? = 20% Ww = ? x ? d = 0. 2x 105 = 21 lbs Total Volume Va Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 37 Borrow Fill Adjustments Borrow Properties: ? m = 110 pcf, ? d =100 & ? = 10% Placed Fill Properties: ? d = 105 pcf & ? = 20% Since borrow ? d =100pcf & fill ? d =105pcf, 105/100 =1. 05 It takes 1. 05 cf of borrow to make 1. 0 cf of fill For 30,000 cy, 30,000 x 1. 05 = 31,500 cy of borrow Total Volume Va Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 38 Borrow Fill Adjustments Borrow Material Properties: Ww = 10 lbs Placed Fill Properties: Ww = 21 lbs Water supplied from borrow in each cf of fill = 10 x 1. 5 = 10. 5 lbs; 21 lbs – 10. 5 = 10. 5 lbs short/1. 05 cf 10. 5lbs/1. 05 cy = 10 lbs of water to be added per cf borrow Total Volume Va Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 39 Proctor: Moisture Density Relationships Establishes the unique relationship of moisture to dry density for each specific soil at a specified compaction energy MOISTURE-DENSITY RELATIONSHIP 108. 0 106. 0 104. 0 D ry D ensity (pcf) 102. 0 100. 0 98. 0 96. 0 94. 0 92. 0 90. 0 88. 0 8. 0 10. 0 12. 0 14. 0 16. 0 18. 0 20. 0 22. 0 24. 0 26. 0 28. 0 Moisture Content (%) 40 Proctor: Moisture Density Relationships • 4” mold 25 blows • 6” mold 56 blows Standard – 5. 5 lb hammer – dropped 12 in – 3 layers Standard: ASTM D-698 AASHTO T-99 Modified: ASTM D-1557 AASHTO T-150 • Modified – 10 lb hammer – dropped 18 in – 5 layers 41 PROCTOR COMPACTION TEST Maximum Dry Density – Highest density for that degree of compactive effort Optimum Moisture Content – Moisture content at which maximum dry density is achieved for 42 that compactive effort Proctor: Moisture Density Relationships MOISTURE-DENSITY RELATIONSHIP 108. 0 106. 0 104. 0 Dry Density (pcf) 102. 0 100. 0 98. 0 96. 0 94. 0 92. 0 90. 0 88. 0 8. 0 10. 0 12. 0 14. 0 16. 0 18. 0 20. 0 22. 0 24. 0 26. 0 28. 0 Moisture Content (%)

What density is required for 95% Compaction? What range of moisture would facilitate achieving 95% compaction? 43 Proctor: Moisture Density Relationships MOISTURE-DENSITY RELATIONSHIP 108. 0 106. 0 104. 0 Dry Density (pcf) 102. 0 100. 0 98. 0 96. 0 94. 0 92. 0 90. 0 88. 0 8. 0 10. 0 12. 0 14. 0 16. 0 18. 0 20. 0 22. 0 24. 0 26. 0 28. 0 Moisture Content (%) 104 x . 95 = 98. 8 pcf A 95% B Range of moisture is within the curve A to B (14 to 24 %) 44 Proctor: Zero Air Voids Line Relationship of density to moisture at saturation for constant specific gravity (SG) Can’t achieve fill in zone right of zero air voids line Z

MOISTURE-DENSITY RELATIONSHIP 108. 0 106. 0 104. 0 Dry Density (pcf) 102. 0 100. 0 98. 0 96. 0 94. 0 92. 0 90. 0 88. 0 8. 0 10. 0 12. 0 14. 0 16. 0 18. 0 20. 0 22. 0 24. 0 26. 0 28. 0 Moisture Content (%) 45 Proctor: Moisture Density Relationships MOISTURE-DENSITY RELATIONSHIP 108. 0 106. 0 104. 0 Dry Density (pcf) 102. 0 100. 0 98. 0 96. 0 94. 0 92. 0 90. 0 88. 0 8. 0 10. 0 12. 0 14. 0 16. 0 18. 0 20. 0 22. 0 24. 0 26. 0 28. 0 Moisture Content (%) If SG = 2. 65 & moisture content is 24% What dry density achieves 100% saturation? A) 100. 0 pcf B) 101. 1 pcf 46 Proctor: Moisture Density Relationships

MOISTURE-DENSITY RELATIONSHIP 108. 0 106. 0 104. 0 Dry Density (pcf) 102. 0 100. 0 98. 0 96. 0 94. 0 92. 0 90. 0 88. 0 8. 0 10. 0 12. 0 14. 0 16. 0 18. 0 20. 0 22. 0 24. 0 26. 0 28. 0 Moisture Content (%) X ?d=SG62. 4/(1+? SG/100) ? d=2. 65×62. 4/(1+24×2. 65/100) ? d=101. 1 pcf Answer is “B” 47 Ref: Peck Hanson & Thornburn Static Head 48 Calculate effective stress at point x Ref: Peck Hanson & Thornburn Saturated Unit Weight ? sat 5’ ? sat = 125 pcf Moist Unit Weight ? M Dry Unit Weight ? Dry 7’ Submerged (buoyant) Unit Weight = ? sat – 62. 4 x 49 Calculate effective stress at point x Ref: Peck Hanson & Thornburn

Total Stress at X 5’ ? sat = 125 pcf = 5 x 62. 4+ 7x 125= 1187psf Pore Pressure at X 7’ = 12 x 62. 4 = 749 psf Effective Stress at X = 1187-749= 438 psf x or (125-62. 4) x 7=438 psf 50 Ref: Peck Hanson & Thornburn Downward Flow Gradient 51 Downward Flow Gradient 3’ Total Stress at X = 5 x 62. 4+ 7x 125= 1187psf Pore Pressure at X ? sat = 125 pcf 7’ = (12-3) x 62. 4 = 562 psf Effective Stress at X = 1187-562 = 625 psf 5’ x or 438 + 3 x 62. 4 = 625psf see previous problem 52 Upward Flow Gradient Ref: Peck Hanson & Thornburn 53 One Dimensional Consolidation ?e/pn 54 Primary Phase Settlement (e log p) ? H = (H x ? )/(1+eo) eo ? H H 55 Consolidation Test Pre-consolidation Pressure Cc = slope of e log p virgin curve est. Cc = 0. 009(LL-10%) Skempton Rebound or recompression curves 56 56 e- l o g p Calculate Compression Index; Cc 1. 50 1. 40 1. 30 Void Ratio (e) 1. 20 1. 10 ksf 0. 1 1 4 8 16 32 (e) 1. 404 1. 404 1. 375 1. 227 1. 08 0. 932 1. 00 0. 90 A) 0. 21 B) 0. 49 57 0. 80 0. 1 1 10 100 Pr essur e ( ksf ) Cc is the slope of the virgin e-log p e- l o g p Cc = -(e1-e2)/log (p1/p2) 1. 50 Cc=-(1. 375-1. 227)/log(4/8) Cc = 0. 49 Answer is “B” ksf 0. 1 1 4 8 16 32 (e) 1. 404 1. 404 1. 375 1. 227 1. 08 0. 932 1. 40 Cc Void Ratio (e) . 30 1. 20 1. 10 1. 00 0. 90 0. 80 0. 1 1 10 100 Pr essur e ( ksf ) 58 Permeability Constant Head Conditions • Q=kiAt • Q= k (h/L)At • k=QL/(Ath) 59 If Q =15cc & t = 30 sec what is the permeability k=QL/(Ath) 10cm 5cm A) 0. 01 cm/sec B) 0. 01×10-2 cm/sec 25cm2 C) 0. 1 cm/sec 60 Constant Head Permeability Calculate k Q =15cc & t = 30 sec • k=QL/(Ath) • k= 15(5)/(25(30)10) • k= 0. 01 cm/sec Answer is “A” 10cm 5cm 25cm2 61 Falling Head Permeability • k=QL/(Ath) (but h varies) • k=2. 3aL/(At) log (h1/h2) • where a = pipette area • h1 = initial head • h2 = final head 62 If t = 30 sec; h1= 30 cm; h2 = 15 cm L= 5 cm; a= 0. cm2; A= 30 cm2; calculate k A) 2. 3×10-3 cm/sec B) 8. 1×10-6 cm/sec C) 7. 7×10-4 cm/sec 63 Falling Head Permeability k=2. 3aL/(At) log (h1/h2) k= 2. 3 (0. 2) 5 /(30×30) log (30/15) k= 7. 7×10-4 cm/sec Answer is “C” 64 •Flow lines & head drop lines must intersect at right angles •All areas must be square •Draw minimum number of lines •Results depend on ratio of Nf/Nd Flow Nets 6ft 2ft 65 Q=kia=kHNf /Nd wt (units = volume/time) w= unit width of section t=time Flow Nets 6ft 66 What flow/day? assume k= 1×10-5 cm/sec =0. 0283 ft/day Q= kH (Nf /Nd) wt Q= 0. 0283x8x(4. 4/8)x1x1 Q= 0. 12 cf/day 2ft Flow Nets ft 67 Check for “quick conditions” pc =2(120)= 240 psf (total stress) Flow Nets Below water level use saturated unit weight for total stress ?= 2(62. 4) = 124. 8 (static pressure) ?? = 1/8(8)(62. 4)= 62. 4 (flow gradient) = 240-(124. 8+62. 4) 2ft 2ft 6ft p’c = pc -(? + ?? ) p’c = 52. 8 psf >0, soil is not quick ?sat=120 pcf 68 Stress Change Influence (1H:2V) For square footing ?? z=Q/(B+z)2 69 If Q= 20 kips, Calculate the vertical stress increase at 7 feet below the footing bottom 5’ 8’ 7’ 70 If Q= 20 kips, Calculate the vertical stress increase at 7 feet below the footing bottom 5’ 8’ ?? z = 0000 (8+7)(5+7) 7’ ?? z = 111 psf 71 Westergaard (layered elastic & inelastic material) If B= 6. 3’ in a square footing with 20 kips load, what is the vertical stress increase at 7’ below the footing bottom? 72 Westergaard Q = 20 kips B = 6. 3’ Z = 7’ ?? z = ? 73 Westergaard 7’/6. 3’ = 1. 1B ?? z = 0. 18 x 20000/6. 32 = 90. 7 psf 74 Boussinesq (homogeneous elastic) Q = 20 kips B = 6. 3’ Z = 7’ ?? z = ? 75 Boussinesq Z/B = 1. 1 ?? z = 0. 3 x 20000/6. 32 = 151 psf 76 Thanks for participating in the PE review course on Soil Mechanics! More questions or comments? You can email me at: [email protected] com 77